Two fractions are such that when 2 is added to thrice the reciprocal of the first fraction we get the second fraction. Also when 4 is subtracted from the denominator of the first fraction we get its numerator. Both the fractions are equal to each other. Let the denominator of the first fraction be d . Write an equation in d .
Note: You don't need to input the equation in quadratic form
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Given Two fractions are such that when 2 is added to thrice the reciprocal of the first fraction we get the second fraction. Also when 4 is subtracted from the denominator of the first fraction we get its numerator. Both the fractions are equal to each other. Let the denominator of the first fraction be d . Write an equation in d .
- Let the fractions be a/d and b/c
- According to the question we get
- 2 + 3d/a = b/c
- Also d – 4 = a
- So a/d = b/c
- Now we have 2 + 3d/a = a/d
- 2a + 3d / a = a/d
- 2(d – 4) + 3d / d – 4 = d – 4 / d (since d – 4 = a)
- (d – 4)^2 = d [2(d – 4) + 3d]
- So d^2 + 16 – 8d = d [2d – 8 + 3d]
- So d^2 + 16 – 8d = 2d^2 – 8d + 3d^2
- 4d^2 = 16
- So d^2 = 4
- Or d = +-2
- Now d – 4 = a
- Taking 2 we get 2 – 4 = a
- Or a = - 2
- Taking – 2 we get – 2 – 4 = a
- Or a = - 6
- So a/d = – 2 / 2 = - 1/1
- Or a/d = - 6 / - 2 = 3 / 1
- Now to find b and c we get
- 2 + 3d / a = b / c
- 2 + 3(2) / - 2 = b/c
- 2 + (-3) / 1 = b/c
- Or b/c = - 1 / 1
- Now taking d = -2 we get
- 2 + 3(-2) / - 6 = b / c
- 2 + 1 = b / c
- Or b / c = 3 / 1
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